#502

Hard

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

IPO

Maximize total capital by selecting up to k projects, based on initial capital and project profits using a greedy strategy.

array greedy sorting heap priority queue

Problem Statement

LeetCode is preparing for its IPO and wants to increase its capital by completing projects. You have an initial capital w and can complete at most k projects from a list. Each project requires a certain amount of capital to start and offers a specific profit once completed. The challenge is to maximize your capital after completing up to k projects, starting with the initial capital and selecting projects based on their profitability and capital requirement.

Given n projects, each with a profit and a capital requirement, your task is to choose up to k projects that maximize the total capital. After completing each project, the profit from that project will increase your available capital, allowing you to start additional projects. The solution should be efficient given the constraints, where both k and n can be large.

Examples

Example 1

Input: k = 2, w = 0, profits = [1,2,3], capital = [0,1,1]

Output: 4

Since your initial capital is 0, you can only start the project indexed 0. After finishing it you will obtain profit 1 and your capital becomes 1. With capital 1, you can either start the project indexed 1 or the project indexed 2. Since you can choose at most 2 projects, you need to finish the project indexed 2 to get the maximum capital. Therefore, output the final maximized capital, which is 0 + 1 + 3 = 4.

Example 2

Input: k = 3, w = 0, profits = [1,2,3], capital = [0,1,2]

Output: 6

Example details omitted.

Constraints

1 <= k <= 105
0 <= w <= 109
n == profits.length
n == capital.length
1 <= n <= 105
0 <= profits[i] <= 104
0 <= capital[i] <= 109

Solution Approach

Greedy Approach

The key to solving this problem is applying a greedy strategy. Always prioritize selecting the project with the highest profit that you can afford based on your current capital. By sorting projects according to their capital requirement and using a max-heap to keep track of the highest profits available, you can efficiently maximize your capital.

Use of Heap

Using a heap (priority queue) allows for efficient retrieval of the most profitable projects that can be started with the available capital. As projects are completed and capital increases, the heap is updated to reflect the new set of possible projects. This ensures you are always selecting the project that maximizes your capital growth.

Invariant Validation

At each step, ensure that your strategy adheres to the invariant of choosing the most profitable project that is affordable with the available capital. By maintaining the heap and ensuring that each decision is based on this invariant, you guarantee the best possible outcome for the problem.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity is dominated by the sorting step and the operations on the heap. Sorting the projects takes O(n log n) time, and each insertion or removal from the heap takes O(log k). Thus, the overall time complexity is O(n log n), which is efficient for large values of n. The space complexity is O(n) due to storing the projects and the heap.

What Interviewers Usually Probe

Candidates who effectively utilize a greedy approach and justify the decision-making process will demonstrate an understanding of the core pattern of the problem.
Look for candidates who efficiently use a heap to manage project selection, as this is key to solving the problem within time limits.
Successful candidates will maintain a clear focus on invariant validation, ensuring that they only choose projects that can be completed with the available capital.

Common Pitfalls or Variants

Common pitfalls

Failing to prioritize the most profitable projects or using an inefficient strategy can lead to suboptimal capital maximization.
Incorrectly managing the heap can cause errors in project selection, leading to incorrect answers or inefficiency.
Not correctly handling the situation where projects cannot be completed due to capital constraints, which might result in unnecessary calculations.

Follow-up variants

Consider scenarios where capital is very large or very small compared to the project costs.
Modify the problem to allow for additional constraints, such as limiting the number of projects from specific categories.
Change the project selection criteria to focus on minimizing risk or maximizing return over time instead of simply maximizing total capital.

FAQ

What is the main strategy for solving the IPO problem?

The IPO problem is solved using a greedy approach combined with a heap to select the most profitable projects that can be completed with the available capital.

Why is a greedy approach appropriate for the IPO problem?

A greedy approach ensures that at every step, we choose the most profitable project that can be completed, which maximizes capital in the long run.

What data structure is crucial for solving the IPO problem efficiently?

A max-heap (priority queue) is crucial for efficiently selecting the most profitable projects that can be completed with the available capital.

How does the heap help in the IPO problem?

The heap helps by maintaining the projects with the highest profit that can be started with the available capital, allowing for efficient project selection and maximization of capital.

What are some common pitfalls when solving the IPO problem?

Common pitfalls include failing to correctly prioritize projects based on profitability, mismanaging the heap, and not handling capital constraints properly.

terminal

Solution

Solution 1

#### Python3

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class Solution:
    def findMaximizedCapital(
        self, k: int, w: int, profits: List[int], capital: List[int]
    ) -> int:
        h1 = [(c, p) for c, p in zip(capital, profits)]
        heapify(h1)
        h2 = []
        while k:
            while h1 and h1[0][0] <= w:
                heappush(h2, -heappop(h1)[1])
            if not h2:
                break
            w -= heappop(h2)
            k -= 1
        return w

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